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Field in the region of the image charge

  1. Jul 28, 2014 #1
    1. The problem statement, all variables and given/known data

    There is an infinite conducting plate in the YZ plane and a point charge q at (d,0,0). I want to find the field for x<0 and the charge distribution on the plate.

    2. Relevant equations



    3. The attempt at a solution

    For the region where x>0, I can use the method of image charges, and hence the potential is:

    [itex] \phi = \frac{q}{4 \pi \varepsilon _0} \left( \frac{1}{\sqrt{(x-d)^2+y^2+z^2}} - \frac{1}{\sqrt{(x+d)^2+y^2+z^2}} \right)[/itex]

    The potential is constant and zero in the plate (x=0), which makes sense. Now, I'd like to calculate the potential on the other side, x<0. I've seen compelling arguments that it should be zero because the plate shields the electric field and that the potential should be zero at infinity... but I'd like to prove it myself, just to be sure. So, I started toying around with the laplacian for x<0 and, using symmetry arguments and 2D solutions, I find that if x<0 then:

    ##\phi = Ax ##

    This is a solution of the laplace equation, it satisfies boundary conditions (##\phi (0,y,z) = 0##), obeys the symmetry of the problem because ##\phi (x,y,z) = \phi (x,-y,z)## and is continuous. So, I'm all out of equations to determine A and prove the potential is 0 on this side. Imposing ##\phi = 0## at infinity would trivially solve it, but I'm not convinced with this because for an isolated plate with uniform charge the potential is definitely not zero at infinity, in fact, it blows up. I believe this has to do with the fact that the charge distribution goes out to infinity, so as long as you have bounded charge distributions you can impose ##\phi \rightarrow 0## as ##|r|\rightarrow \infty##. I know this is the case in this particular problem, but because I saw the solution and I've seen the resulting charge distribution. However, I need the potential to calculate the charge distribution and I need the charge distribution to use the argument to calculate the potential. I'm stuck in an argumental loop!

    Does anyone know any way (that doesn't involve ##\phi \rightarrow 0## as ##|r|\rightarrow \infty##) to prove that the constant A must be zero? If there isn't any other way, when and why can you use it? Is it physically necessary??? Because taking any A will actually solve the problem (it verifies the laplace equation and the boundary condition), the only condition that doesn't check out is that of the potential becoming zero at infinity.

    I think this workflow might work, aided by the uniqueness of solutions. Maybe somebody can give me some input:
    1) Assume there isn't any charge at infinity
    2) Solve the problem under this assumption, that means, using that the potential is zero at infinity
    3) Calculate the charge distribution and verify the hypothesis was correct.

    If I were to choose the different hypothesis:

    1) Assume there is charge at infinity
    2) Then the potential does not go to zero and A is a non zero constant
    3) Calculate the charge distribution and verify the hypothesis (the calculated charge distribution should be somewhat different, but still zero at infinity, hence, the hypothesis was wrong).

    Note that the only hypothesis that doesn't check out is that of there being charge at infinity, continuity and boundary conditions still work just fine.

    Finally, I'd like to add a few questions myself. First, is the field zero because the charge distribution arranges itself in such a way that it cancels out the point charge? Second, if the plate is a conductor and the charge arises from the polarization produced by the point charge, shouldn't there be an equal and opposite charge induced somewhere else? (Because the conductor is neutral overall).

    I'm sorry if I wasn't very clear, I tried to illustrate my reasoning and it was kind of a mess :D.

    Thank you guys in advance
     
  2. jcsd
  3. Jul 30, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You can not make an infinite plate. If you have a big but finite one, the plate is usually said connected to the ground (earthed). There is no problem then, the side facing the charge q has a total induced charge -q, coming from the ground, and no charge is present on the other side: the electric field is zero for x<0.
    In case the plate is not earthed, the induced charge comes from the metal, so a total of opposite charge, q, appears on the other side. As the field is zero inside the plate, the charges on the opposite surfaces do not feel each other. The charge distribution on the opposite face is homogeneous and the magnitude of the electric field is σ/ε0. If the potential is zero at the plate, it will blow up at infinity.

    But: At very far away, the size of the plate is small with respect to the distance from it. The edge effects start to dominate the field, and the electric field is not homogeneous at x<0 any more.

    There is also the possible argument that an infinite plate extends to infinity so it is connected to infinity, as if it was earthed. It has the same potential as the infinity opposite to it. So there is no electric field in between. Or it gains the -q charge from infinity.

    But it is the best if you say that the plate is grounded.

    ehild
     
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